Question

A 250-V, 10-A dynamometer type watt meter has resistance of current and potential coils of 0.5 and 12,500 ohms respectively. Find the percentage error due to each of the two methods of connection when unity p.f. loads at 250 volts are of (a) 4A (b) 12 A.

Neglect the error due to the inductance of pressure coil.

Answer 1

(a) When I = 4 A

(i) Consider the type of connection

Power loss in current coil of watt meter = I2r = 42 × 0.5 = 8 W

Load power = 250 × 4 × 1 = 1000 W;

Watt meter reading = 1008 W

∴ percentage error = (8/1008) × 100 = 0.794%

(ii) Power loss in pressure coil resistance = V2/R

= 2502 /12,500 = 5 W

∴ Percentage error = 5 × 100/1005 = 0.497 %

(b) When I = 12A

(i) Power loss in current coil = 122 × 0.5 = 72 W

Load power = 250 × 12 × 1 = 3000 W ;

watt meter reading = 3072 W

∴ percentage error = 72 × 100/3072 = 2.34 %

(ii) Power loss in the resistance of pressure coil is

2502 /12,500 = 5 W

∴ percentage error = 5 × 100/3005 = 0.166 %